1.3,7,10,11,12,17,? find out next number?
Sol: 3,7,10,11,12,17
sum of alternates -1 equal to next alternate number.
3+10-1=12
7+11-1=17
so the next number will be 10+12-1=21
2. Cp of 4 calculators and 2 pencil is is 6200 what is the cost of ten calculators and five pencils
Sol: 4C+2P=6200–>2C+P=3100
SO 10C+5P=5(2C+P)=5(3100)=15,500 RS
3.12 men can complete work in 6 dayswhereas 10 men and 21 women take 3 days to finish the same work .in how many days can 12 women alone complete
Sol: 10 men’s, 1 day work=10/(12*6)=5/36
If 21 women’s, 1 day work= 21/W ,then
3[(5/36)+(21/W)]= 1 ,On solving, W=108
So, 12 women can complete the work in 108/12=9 days
4. 27^18/14 find the remainder value?
Sol: any number of the form (a*x-1)^n /a the remainder will be +1 if the power n is even. and the remainder will be -1 or (a-1) if the power is odd. According to this the remainder will be 1
5. Probability of getting sum of odd numbers in two throw of a dice?
Sol: odd numbers 3,5,7,9,11(between 2(min sum)-12(max sum))
cases:-
3-(1,2),(2,1)
5-(3,2),(2,3),(4,1),(1,4)
7-(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)
9-(3,6),(4,5),(5,4),(6,3)
11-(5,6),(6,5)
total cases=18
therefore probability=18/36=>1/2
6. if log(p+q)(p-q)= -1 then find the value of: log(p+q)(P2-q2) (this p square – q square)
Sol: log(p+q)(p-q)=-1
log(p^2-q^2)= log(1/10)adding both side log(p+q)
then,

Number of 2s in 77! is: = 38 + 19 + 9 + 4 + 2 + 1 = 73
So number of 2^4 will be: = 73%4=18
Number of 3s in 77! is:= 25 + 8 + 2 = 35
So number of 3^2 will be: = 17
number of 5s will be:= 15 + 3 = 18
Hence the number of 720s that we can obtain by analyzng d minimm value 17…
10. Some persons can do a piece of work in 12 days. Two times the number of such persons will do half of that work in:
Sol: let no of persons can do x work
1 day work=1/12
now x/2=1/6
6x=2
x=1/3
ans:3 days
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11. If log102 = 0.3010, what is the number of digits in 264
Sol:It is always true that
log(base 10)(a three digit no)= gives a characteristics value 2″.and to find the no of digits we use the formula “characteristics value+1″…so 264 is a three digit no and its characteristics value for log(base 10) 264 is 2 and after decimal there is xyz anything so according to formula it is 2 + 1 = 3…
12. log y 1369y=3 then what is the value of y?
Sol: log y raised to the power of 3;
y^3 = 1369y
y^2 = 1369
y =37.
13. What is the remainder when 17^23 is divided by 16?
Option 1 : 0 Option 2 : 1 Option 3 : 2 Option 4 : 3
Sol: 17^23 /16
(16+1)^23 = 16^23 +1^23 +16*(…….)
When divided by 16 it leaves 1 as reminder
14. antilog10^100
Sol: antilogx=10^x(for base 10) so ans is 10^(10^100)

15. Four bells begin to toll together and then each one at intervals of 6 s, 7 s, 8 s and 9 s respectively. The
number of times they will toll together in the next 2 hr is:
Option 1 : 14 times Option 2 : 15 times Option 3 : 13 times Option 4 : 11 times
Sol: given each interval time is 6s.7s,8s,9s.
Now all together toll is nothing but lcm of all individuals i.e., lcm of6,7,8,9=504s
convert 2 hours in secondsi.e., 2hrs=2*60*60=7200s
now no.of times=total time in seconds/together lcm
i.e,7200/504=14
16. The students are in the ratio 2:3:5.if 20 students are increased in each batch the ratio changes to 4:3:7
The total number of students in the three batches before the increase was
Sol: the ratio is 2:3:5
let students 2x,3x,5x.
After 20 student are increased in each batch
now student will be 2x+20,3x+20,5x+20
now ratio is 4:3:7
now student will be 4x,3x,7x
2x+20=4x
x=10
students before increases : 20,30,50
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17. sum of money doubles itself in 9 years, in how many years it will become 8 times itself?
Sol: Given, The Sum Of Money Doubles Itself In 9years
Initially Let The Sum Be X
After 9 Years It Becomes 2X(Now This Becomes The Initial For Next 9y)
After 18 Years It Becomes 4X(Now This Becomes The Initial For Next 9y)
After 27 Years It Becomes 8X.
So It Takes 27 Years To Get 8times Of The Initial Amount Invested.
18. What is the smallest four-digit number which when divided by 6, leaves a remainder of 5 and when
divided by 5 leaves a remainder of 3?
Option 1 : 1043 Option 2 : 1073 Option 3 : 1103 Option 4 : None of
these
Sol: ans is none of these because :
let us assume the smallest 4 digit number be 1000 if we divide it with 6 we get remainder 4 so to get a rem of 5 add 1 to it => 1001.
Then the general form of a number is 1001+6k for every positive integer value of k it always yields rem 5 when divided by 6,then by trail and error if we take k=2 then number is 1013 which when divided by 5 gives a rem of 3 so the right ans is 1013 which is none of these from options
19. A, B, C started a business with their investments in the ratio 1:3:5. After 4 months, A invested the same amount as before and B as well as C withdrew half of their investments. The ratio of their profits at the end of the year is:
Sol: Let their initial investments be x, 3x and 5x respectively. Then,
A : B : C = (x * 4 + 2x * 8) : (3x * 4 + 3x/2 * 8 ) : (5x * 4 + 5x/2 * 8)
= 20x : 24x : 40x = 5 : 6 : 10.
20. There are 10 yes or no questions. How many ways can these be answered
Sol: Each question can be answered either yes or no, so every question will have 2 possibilities
therefore, for 10 questions, 20 possibilities(2+2+2+….10(times)
21. If (a^4 – 2a^2b^2 + b^4)^x-1 = (a-b)^2x (a+b)^-2, then x equals to:
Sol: Taking log on both sides,
=>(x-1)log(a2-b2)^2=log[(a-b)^2x(a+b)^-2]
=>2(x-1)log(a2-b2)=2xlog(a-b)-2log(a+b)
Cancelling 2 on both sides,and expanding log(a2-b2) into log [(a+b)(a-b)]=>log(a+b)+log(a-b),
=>(x-1)[log(a+b)+log(a-b)]=xlog(a-b)-log(a+b)
=>xlog(a+b)-log(a-b)-log(a+b)+xlog(a-b)=xlog(a-b)-log(a+b)
=>xlog(a+b)=log(a-b)
x=log(a-b)/log(a+b).
22. In an examination, 70% of students passed in physics, 65% in chemistry, 27% failed in both subjects. The percentage of students who passed is: Op 1: 66% Op 2: 62% Op 3: 69% Op 4: None of these
Sol: pass % in physics n(P)=70%
pass %in chemistry n(c)=65%
fail % in both=27%
pass %in any one or both=n(pUc)=73%
n(p intersection c)=n(p)+n(C)-n(pUc)=70+65-73=62%
23. If the simple interest on a sum at 4% per annum for 2 years is Rs. 80, then the compound interest on the same sum for the same period is: Op 1: Rs. 86.80 Op 2: Rs. 86.10 Op 3: Rs. 88.65 Op 4: Rs. 81.60
Sol: Amount from simple interest is 1000.then calculate the amount from this formula
a=p(1+r/100)^n.
so,1000(1+4/100)^2 =1081.6.
so, interest is 81.6.
24.Prabodh bought 30 kg of rice at the rate of Rs. 8.50 per kg and 20 kg of rice at the rate of Rs. 9.00 per kg. He mixed the two. At what price (App.) per kg should he sell the mixture in order to get 20% profit? Op 1: Rs. 9.50 Op 2: Rs. 8.50 Op 3: Rs. 10.50 Op 4: Rs. 12.00
Sol: 30 Kg->8.50*30=255
20kg -> 9*20 =180
so
50Kg=435; 1kg=435/50=8.7.

So CP of 1kg=8.7
now 20% profit it means -> (20/100)*8.7=1.74.

so the final ans is 8.7+1.74=”10.44″
25. Mohan walks a certain distance and rides back in 6 hours and 15 minutes. If he walks both ways he takes 7 hours and 45 minutes. If Mohan rides both ways the time which he will take will be: Op 1: 4 hours Op 2: 19/4 hours Op 3: 9/2 hours Op 4: 17/4 hours Op 5: None of these
Sol: W+R=375 minutes(6 hours 15 minutes)
2W=465 minutes(& hours 45 minutes)
2R=?
2(W+R)=375*2=750
2R=750-465=285=19/4
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26. In an examination 10 questions are to be answered choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can 10 questions be answered ? Op 1: 212 Op 2: 266 Op 3: 272 Op 4: 312
Sol: Here total question 10 will be answered so there are 3 ways .i know A has 6 nd B has 7 so
A. B
4. 6
5. 5
6. 4
so total no of ways is 6c4*7c6+6c5*7c5+6c6*7c4=266 so its the ans.
27.A boy move 6 m in west then he turn towards south and move 20 m then turn towards east and move 12 m again move toward north and move 12 m . How much dist he is away from his starting point
Sol: it forms a right angle triange so
hypotenuse square= side_1 square + side_2 square
one side it is 6 n the other it is 8 so
6^2 + 8^2=100 so ans 10
28. synonym of OBTRUSIVE
Sol: conspicuous, obvious,unmistacable
29. hcf of 3.68 & 5.35

Reason being 11 coins are picked randomly and placed in a box, and now the coin is picked from those 11..

So how can we know that between those 11 coins how may are 1Rs coins and how many 50 paisa… so “can not determine”
PLEASE REFER ONCE!!!
49. . A, B and C are three students who attend the same tutorial classes. If the probability that on a particular day exactly one out of A and B attends the class is 7/10; exactly one out of B and C attends is 4/10; exactly one out of C and A attends is 7/10. I
Sol: Probability(at least one attending ) = 1- Probability(none attending)
Let the Probability of A,B,C attending the class be a,b,c
So not attending will be 1-a,1-b,1-c
Exactly one of A,B
a(1-b) + b(1-a) = 7/10
a+b -2ab =7/10
B,C
b(1-c)+c(1-b) = 4/10
b+c -2bc =4/10
C,A
a(1-c) + c(1-a) =7/10
c+a -2ac=7/10
Add all 3 u get
2(a+b+c) – 2(ab+bc+ca) = 18/10
a+b+c -ab -bc-ca =9/10
P(atleast one ) = 1 – P(none)
1 – [ (1-a)(1-b)(1-c)]
1- [ 1 -a -b-c+ab+bc+ca -abc]
1-[1 – (9/10 + 9/100)]
= 99/100

50. A box contains 10 balls numbered 1 through 10. Anuj, Anisha and Amit pick a ball each, one after the other, each time replacing the ball. What is the probability that Anuj picks a ball numbered less than that picked by Anisha, who in turn picks a lesser n Op 1: 3/25 Op 2: 1/6 Op 3: 4/25 Op 4: 81/400
Sol: If Amit picks up 10 and Anisha picks up 9 then

You either add them or apply the expression n(n+1)(n+2)/6 which is the sum of triangular numbers =8(9)(10)/6 = 120

Probability = 120/(10 x 10 x 10) = 3/25
51. A, B, C, D and E play the following game. Each person picks one card from cards numbered 1 through 10. The person who picks the greatest numbered card loses and is out of the game. Now the remaining four return their cards to the pack and draw again, and Op 1: 3/14 Op 2: 4/17
Op 3: 1/5 Op 4: 5/24
Sol:
52. a buy clips at 12 for R.s. 60 .How many clips should he sellfor Rs. 60 to earn a profit of 20% ?

a>5
b>8
c>6
d>10
Sol: no.of clips =60/12=5
let x clip should be sell to earn a 20% profit.
in this contrast,
x clip sell for rs 60
1 clip sell for rs:60/x;
proffit=(60/x-5)/5*100=20/100;
x=10(ans)
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53. An article was sold for Rs. 2770. Had it been sold for Rs. 3000 there would have been an additional gain of 10%. Cost Price of the article is:
Op 1: Rs. 2100
Op 2: Rs. 2200
Op 3: Rs. 2300
Op 4: Rs. 2400
Sol: ans : 3
given selling price is = 2770
he said if we sell it for 3000 there would be a 10% more gain
3000-2770=230
because of this Rs.230 he can gain 10% more
from profit percentage formula
230*100/ cost price =10
from the above equation cost price is 2300
54. The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target at least three times is:
Op 1: 291/364
Op 2: 371/464
Op 3: 471/502
Op 4: 459/512
Sol: ans: 4
hitting the target at least 3 times means it can be greater than 3 also i.e.3,4,5
in 5 chances hitting target by 3 times is
5c3*(3/4)^3*(1/4)^2 = 10*27/1024= 270/1024
probability of hitting by 4 times is
5c4*(3/4)^4(1/4)^1=5*81/1024= 405/1024
probability of hitting 5 times is
5c5*(3/4)^5 = 243/1024
total is (270+405+243)/1024= 918/1024
=459/512
55. A 5-digit number is formed by the digits 1,2,3,4 and 5 without repetition. What is the probability that the number formed is a multiple of 4?
Op 1: 1/4
Op 2: 1/5
Op 3: 2/5
Op 4: 1/120
Op 5: 4
Sol: last two digits must be,
***12
***24
***32
***52
case1) ***12
rest three digits can be filled in 3*2*1 or 6 ways
similarly,
case2. ***24
6 ways
case3. ***32
6ways
case4. ***52
6 ways
total required ways = 6*4 = 24
total arrangements = 5*4*3*2*1 = 120

probability = 24/120 = 1/5
56. In how many ways can a number 6084 be written as a product of two different factors ?
Op 1: 27
Op 2: 26
Op 3: 13
Op 4: 14
Sol: First find the prime factors of 6084
2*3042
2*2*1521
2*2*3*507
2*2*3*3*169
2*2*3*3*13*13

2^2*3^2*13^2

Number of factors = (2+1)*(2+1)*(2+1) = 3*3*3 = 27
Using each of these factors, we can write 6084 as a product of 2 factors.
Half of these will remain same.
Let us take a simple example, 6
number of factors = 4
1,2,3,6
can be written as
1*6
2*3
We cant repeat 6*1 and 3*2 as they are already taken.
Hence only 2 ways.
Another example: take 9.
Number of factors = 3
1,3,9
can be written as
1*9 only.
Hence only 1 way.
So it turnsout ot be number of factors/2 if number of factors is even and
(number of factors-1)/2 if number of factors is odd.
Therefore: 27-1/2 = 26/2 = 13
Hence pick Op 3.
57.A lady gives dinner party to five guests to be selected from 9 friends .The number of ways of forming the party of 5,given that two of the friends will not attend the party together is
Sol: No of guests to be invited=5
Therefore,
No of ways forming the party=
=(9-2)c5*2c0+(9-2)c4*2c1
=7c5*1+7c4*2
=91
58. There are 5 letters and five addressed envelops. the number of ways in which all the letters can be put in wrong envelops is
Sol: We have N letters and N envelopes. The Letters can be put in the N envelopes in N! ways . We want to count the Number of “Derangements” ( The no. of ways that no letter goes into right envelope ).
N!( 1- 1/1! + 1/2! – 1/3! +………………+(-1)n 1/n! ) (this the the formula).
Here N = 5.
So When We put N = 5 in Formula we get 44 ans.
59. A five -digit number divisible by 3 is to be formed using numerals 0,1,2,3,4 and 5 without repetition. The total number of ways this can be done is
Sol: Total 5-digit Number formed by 0,1,2,3,4,5 (without repetition) are 5*5*4*3*2 = 600.
Number divisible by 3 means sum is divisible by 3.
so case 1. 5+4+3+2+1 = 15(exclude 0).
possibilities are = 5*4*3*2*1 = 120
case 2. 5+0+4+2+1 = 12(exclude 3).
possibilities are = 4*4*3*2*1 = 96.
so total answer = case1 + case2 = 216
60. Mark price of a good is 45 Rs. If seller sells it at 42 Rs as discount price and also want 5 % profit then what will be cost price?
Sol: sp=gain% * cp

42=105% * cp (5% profit)

cp= (42*100)/105 = 40
61. How many 4 digit even no. is possible by 1,2,3,4 if no one is repeated?
Sol: unit place can be 2 or 4 only (2 ways) rest 3 places can be filled in 3P3 = 3! = 6 ways total no. = 2*6 = 12
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62. log3 9-log4 256+log5 125=?
Sol: log3 9 – log4 256 + log5 125
= log3 3^2 -log4 4^4 + log5 5^3
= 2*log3 3 – 4*log4 4 + 3*log5 5
= 2 – 4 + 3 [log x (x) = 1]
= 1
63. If a=2 & b=1 then log(a+b)(a^2-b^2)=?
Sol: if(a+b) is base then
log(a+b)(a^2-b^2)
= log(2+1) (2^2-1^2)
= log3 (3)
= 1
64. a coin is tossed 3 times by raju.what is probability that raju win all three time.
Sol: every time event is independent
prob. of win 0.5 for one time
for three times 0.5*0.5*0.5 = 0.125
65. If there are 5 different roads to go into a city then no. of ways to go and back to home?
a)5
b)10
c)25
d)20
Sol: if one goes using 1st road,there are 5 roads to come back…..so 5*5,25 is the ans
66. Find next no. in sequence
8, 12, 24, 60, ?
Sol: every times difference increases three times.
4,12,36..,108
so the nxt number should be 60+108=168
67. probability of finding 9 of hearts from deck of 52 cards ?
Sol: there is only 1,9 of heart is present in a deck of 52 cards.
so probability of finding 9 of heart=1/52
68. log rootover(6) 1296= ?
Sol: log rootover(6) 1296

= log 6^.5 (6^4)
= 1/.5 * 4 * log 6(6)
= 2*4*1
=8
69. A and B start together from the same point on a circular track and walk in the same direction till they both again arrive together at the starting point. A completes one circle in 224 s and B in 364 s. How many times will A have passed B?
Sol: LCM of 224,364=2912
so, A does 13 circles while B does 8 in 2912s.
Thus A crosses B 13 times.
70. Hemant and Ajay start a two-length swimming race at the same moment but from opposite ends of the pool. They swim in lane and at uniform speed, but Hemant is faster than Ajay.They first pass at a point 18.5 m from the deep end and having completed one length, each one is allowed to rest on the edge for exactly 45 seconds.After setting off on the return length, the swimmers pass for the second time just 10.5 m from the shallow end.How long is the pool?

Options:
a.55.5m, b.45m, c.66m, d.49m.
Sol: When they are meeting for the first time, sum of distance travelled by both swimmers is = d, if ‘d’ is the required length of the pool. Now when they are meeting for the second time (i.e. in their reverse journey), sum of the distance travelled by both swimmers = 3d.

If we exclude the waiting time, then time taken for first meet and second meet is in the ratio 1 : 3. Also distance travelled by each swimmer will be in the ratio 1 : 3 (as both are moving with constant but distinct speeds). So equating the ratio of distance travelled by one of the swimmer with the ratio 1 : 3, we get

railway accidents.
Assumptions:
I. The government has enough funds to meet the expenses due for compensation.
II. There may be reduction in incidents of railway accidents in near future.
Op 1: Only Assumption I is implicit.
Op 2: Only Assumption II is implicit.
Op 3: Either Assumption I or II is implicit.
Op 4: Neither Assumption I nor II is implicit.
Op 5: Both Assumptions I and II are implicit.
Sol:C orrect Op : 1
106. Ques. Statement:
‘Please do not wait for me, I may be late, start taking lunch as soon as the guests arrive.’ – a message from a Director

of a Company to his Office managers.
Assumptions:
I. Keeping guests waiting is not desirable.
II.Lunch may not be ready in time.
Op 1: Only Assumption I is implicit.
Op 2: Only Assumption II is implicit.
Op 3: Either Assumption I or II is implicit.
Op 4: Neither Assumption I nor II is implicit.
Op 5: Both Assumptions I and II are implicit.
Sol: Correct Op : 1
107. Statement:
The government has instructed all the premier institutes offering professional courses to reduce the fees by 50 percent

and increase the number of students.
Assumptions:
I. These institutes may be able to continue providing quality education with less fees and more students.
II. The institutes may continue charging more fees to provide quality education.
Op 1: Only Assumption I is implicit.
Op 2: Only Assumption II is implicit.
Op 3: Either Assumption I or II is implicit.
Op 4: Neither Assumption I nor II is implicit.
Op 5: Both Assumptions I and II are implicit.
Sol: Correct Op : 5
108. Statement:
The railway authority has decided to introduce two additional super-fast trains between Cities ‘A’ and ‘B’ during the
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vacation time.
Assumptions:
I. All the passengers who desire to travel during vacation time will get a train ticket.
II. All other modes of transport between cities ‘A’ and ‘B’ are already overstretched.
Op 1: Only Assumption I is implicit.
Op 2: Only Assumption II is implicit.
Op 3: Either Assumption I or II is implicit.
Op 4: Neither Assumption I nor II is implicit.
Op 5: Both Assumptions I and II are implicit
Sol: Correct Op : 2